MCQ
$\triangle\text{ABC}$ is a right triangle right angled at $C$, then the value of $\text{cosec}^2\text{A}-\text{sec}^2\text{B}$ is:
- ✓$0$
- B$-1$
- C$2$
- D$1$
✓
Answer
Correct option: A.
$0$
In $\triangle \text{ABC}$
$\angle \text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\angle\text{B}=90^\circ-\angle\text{A}$
$\text{csc}^2\text{A}-\text{sec}^2\text{B}$
$\text{csc}^2\text{A}-\text{sec}^2(90^\circ-\angle\text{A})$
$(\therefore\text{sec}(90^\circ-\angle\text{A})=\text{csc A})$
$\text{csc}^2\text{A}-\text{csc}^2\text{A}$
$=0$
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